Wolfgang Jeltsch wrote:
>>This seems to suggest:
>>>> Add a == exists (add :: a -> a -> a)
>>>>>>Doesn't "exists" normally quantify over types and not over values?
>>It is quantifying over types, it is saying there exists a type "a -> a
-> a" that has
at least one value we will call "add"...
I think the important point is that the existential is a pair of (proof,
proposition)
which through curry-howard-isomorphism is (value in set, set). Here we
are saying that
there is a set of "functions" with the type "a -> a -> a" ... for the
existential to be satisfied
there must be one called "add". Consider this as an assumption placed on
the environment
by the function.
Keean.